Question

Exercise 2: (7marks) um voltage three-phase line delivering St5MVA with cosp = 0.928 ogging. Teeds a balanced three-phase lood that consumes an active power and imposes a power factor of 0.938 lagging (Figure 2) Each line wire has resistance R=2.43 and inductance L - 11.2 MH. The effective (RMS) phase-to-phase voltage at the starting of the line is 0o = 21.0 kV The frequency of the voltage is 50 Hz. The purpose of the problem is to calculate the drop voltage due to the line. R Uo IR P = ? W DRL COS 0,938 Figure 2 1-Calculate the active power consumed by the load. Y 6996,53 2-Calculate the voltage UA at the arrival line of the load. SX V2X 3-Deduce the drop voltage due to the line.

Answer 1

Up= LOKV S=sinua 0.928 of lig fi s= 5/+91-87 MVA S=BVI* =N3x 21*2* = 5421-87° D2 = 0.137464(+21.87° E = 0.1374 6 4 621 XL= 21762 = = 20 X 50 X 11.28103 3:5185 r . 2,43 3.5185 mom 4= 27 - (-43 +j3.5185) *(0.137464 (-21870) VA= 11638714-59732 kV (VA) line = 20.15882-1-59732 KN Total power = 73 (VA); Pante =V3x{20.1588 1–1-59732 )*(0.13 7464211870

n Total Power = (415023 +1 1.66304) MVA. 1) Active power consumed by load = 4.5023 MW 20 UA= 20.1588 KW 3) voltage drop - 21 - 20:1588 (-.59732° = 11018143 / 33.498° KV